Problem: Segment $AB$ has midpoint $C$, and segment $BC$ has midpoint $D$. Semi-circles are constructed with diameters $\overline{AB}$ and $\overline{BC}$ to form the entire region shown. Segment $CP$ splits the region into two sections of equal area. What is the degree measure of angle $ACP$? Express your answer as a decimal to the nearest tenth.

[asy]
draw((0,0)--10dir(180),linewidth(2));
draw((0,0)--10dir(67.5),linewidth(2));
draw((0,0)--10dir(0),dashed);

draw(10dir(180)..10dir(90)..10dir(0),linewidth(2));

draw((5,0)+5dir(180)..(5,0)+5dir(-90)..(5,0)+5dir(0),linewidth(2));

dot((0,0));
dot((5,0));

label("A",10dir(180),W);
label("B",10dir(0),E);
label("C",(0,0),SW);
label("D",5dir(0),NE);
label("P",10dir(67.5),NE);
[/asy]
The semi-circle with diameter BC has radius $\frac{1}{2}$ that of the semi-circle with diameter AB, and thus, has $\frac{1}{4}$ of the area.  (Area of a circle $= \pi \times r^2$ - thus, if $r$ is half as large, that will be squared in the process).  Therefore, the sum of their areas represents $\frac{5}{8}$ of a circle with diameter AB, and since the line CP splits this area exactly in half, that area would be $\frac{5}{16}$ of a circle with diameter AB.  Therefore, the degree measure of that sector is $360 \times \frac{5}{16} = \boxed{112.5}$